- #1

- 307

- 19

- #1

- 307

- 19

- #2

mfb

Mentor

- 35,264

- 11,519

There are ellipses where both focal points are at the same place, but this special case has a different name which is usually preferred.

- #3

- 16,738

- 7,425

Your thinking makes no sense. You may be thinking of a parabola. If you are really thinking of an ellipse, then you just need to study them more.

- #4

- 307

- 19

so please explain me why ellipse has two focus

- #5

- 16,738

- 7,425

If you spend any time studying the math of the ellipse it is very obvious why it has to focii. It's all there in the math and it's not hard. I'm puzzled by your confusion. Have you used the equation of an ellipse to draw one?so please explain me why ellipse has two focus

- #6

- 307

- 19

- #7

- 16,738

- 7,425

Ah, I see your confusion now. Just keep in mind that it is sometimes true that you can use different forms to describe the same thing but if they are indeed describing the same thing, then they have to be equivalent.

- #8

mfb

Mentor

- 35,264

- 11,519

- #9

- 307

- 19

Then we have SA = e AZ ...... (1)

And SA' = e A'Z ...... (2)

.·. A and A' lie on the ellipse.

Let AA' = 2a and take O the midpoint of AA' as origin. Let P(x, y) be any point on the ellipse referred to OA and OB as co-ordinate axis.

Then from figure it is evident that

AS = AO - OS = a - OS

AZ = OZ - OA = OZ - a

A'S = A'O + OS = a + OS

A'Z = OZ + OA' = OZ + a

Substituting these values in (1) and (2), we have

a - OS = e (OZ - a) ...... (3)

a + OS = e (OZ + a) ...... (4)

Adding (3) and (4), we get

2a = 2 e OZ

Or OZ = a/e ...... (5)

Subtracting (3) from (4), we get

2 OS = 2ae => OS = ae ...... (6)

.·. The directrix MZ is x = OZ = a/e and the co-ordinate of the focus S are (OS, 0) i.e. (ae, 0). Now as P(x, y) lies on the ellipse.

So we get

SP = e PM or SP2 = e2 PM2

(x - ae)2 + y2 = e2 [OZ - x co-ordinate of P]2

=> (x - ae)2 + y2 = e2 [a/e - x]2 = (a - ex)2 ...... (7)

=> x2 + a2e2 - 2axe + y2 = a2 + e2x2 - 2aex

or x2/a2 + y2/a2(1-e2) = 1 [Dividing each term by a2 (1 - e2)]

or x2/a2 + y2/b2 = 1 where b2 = a2 (1 - e2)

in this proof they have used only one focus, i know that from both the focus equation will be same. but this made me confused to treat ellipse having only one true focus.

- #10

- 307

- 19

from this proof how can we say that ellipse has two foci

- #11

mfb

Mentor

- 35,264

- 11,519

- #12

- 307

- 19

yes right

- #13

symbolipoint

Homework Helper

Education Advisor

Gold Member

- 6,115

- 1,161

The requirement of two foci comes from the definition. The set of points in a plane whose sum of distance from two fixed points are equal. Said better in the first paragraph of this article: https://en.wikipedia.org/wiki/Derivation_of_the_Cartesian_form_for_an_ellipse [Broken]

Last edited by a moderator:

- #14

SammyS

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

- 11,358

- 1,031

Now that this topic has be discussed rather well, we can state that the name mfb refers to is "circle".

There are ellipses where both focal points are at the same place, but this special case has a different name which is usually preferred.

- Last Post

- Replies
- 6

- Views
- 1K

- Replies
- 2

- Views
- 2K

- Replies
- 2

- Views
- 302

- Replies
- 8

- Views
- 7K

- Replies
- 2

- Views
- 10K

- Last Post

- Replies
- 4

- Views
- 850

- Replies
- 10

- Views
- 5K

- Replies
- 3

- Views
- 4K

- Last Post

- Replies
- 18

- Views
- 5K

- Last Post

- Replies
- 6

- Views
- 9K